Problem: Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11

Solution (in CPP):

Solution:
#include<iostream>
#include<iomanip>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<sstream>
using namespace std;
 
#define P(p) cout<<(p)<<endl
#define rep(i,m,n) for(int i = (m); i < (int)(n); i++)
#define rrep(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define vsort(v) sort(v.begin(), v.end());
#define rvsort(v) sort(v.begin(), v.end(),greater<int>());
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define ret return
#define lb(v,n) lower_bound(v.begin(),v.end(),n)
#define ub(v,n) upper_bound(v.begin(),v.end(),n)
#define mae(v) max_element(v.begin(),v.end())
#define mie(v) min_element(v.begin(),v.end())
#define INF 0x7FFFFFFF
#define mod 1000000007
#define fr first
#define sc second
typedef long long ll;
////////////////////////////////////////////////////////////

typedef struct RollingHash {
    typedef long long int_type;
    typedef pair<int_type,int_type> hash_type;

    int_type base1;
    int_type base2;
    int_type mod1;
    int_type mod2;
    vector<int_type> hash1;
    vector<int_type> hash2;
    vector<int_type> pow1;
    vector<int_type> pow2;

    RollingHash() : base1(1009), base2(1007), mod1(1000000007), mod2(1000000009) {}

    void init(const string &s) {
        int n = s.size();

        hash1.assign(n+1,0);
        hash2.assign(n+1,0);
        pow1.assign(n+1,1);
        pow2.assign(n+1,1);

        for(int i=0;i<n;i++) {
            hash1[i+1] = (hash1[i]+s[i]) * base1 % mod1;
            hash2[i+1] = (hash2[i]+s[i]) * base2 % mod2;
            pow1[i+1] = pow1[i] * base1 % mod1;
            pow2[i+1] = pow2[i] * base2 % mod2;
        }
    }

    hash_type get(int l,int r) {
        int_type t1 = ((hash1[r] - hash1[l] * pow1[r-l]) % mod1 + mod1) % mod1;
        int_type t2 = ((hash2[r] - hash2[l] * pow2[r-l]) % mod2 + mod2) % mod2;
        return make_pair(t1, t2);
    }

    RollingHash::hash_type concat(hash_type h1, hash_type h2, int h2_len) {
        return make_pair((h1.fr*pow1[h2_len]+h2.fr)%mod1, (h1.sc*pow2[h2_len]+h2.sc)%mod2);
    }

}rh;

int main(){
	int n,m;
	string s;
	cin >> n >> m >> s;
	
	rh r;
	r.init(s);
	
	typedef unsigned long long ull;
	vector<pair<ull,ull>> ans;
	
	ll rr = 1 , l = 0;
	rep(i,0,m){
		string in;
		cin >> in;

		if( in == "R++") rr++;
		else if(in == "R--") rr--;
		else if(in == "L++") l++;
		else l--;
		
		ans.push_back( r.get(l,rr) );	
	}

	vsort(ans);
	cout << unique( ans.begin() , ans.end() ) - ans.begin() << endl;

	ret 0;
}